pandas 中的逐元素 XOR
Element-wise XOR in pandas
我知道逻辑与是&，逻辑或是|在 Pandas 系列中，但我一直在寻找按元素逻辑异或。我想我可以用 AND 和 OR 来表达它，但如果可以的话，我更愿意使用 XOR。
谢谢！
Python异或：a ^ b
Numpy logical XOR: np.logical_xor(a,b)
测试性能 - 结果相等：
1.大小为 10000
的随机布尔序列
In [7]: a = np.random.choice([True, False], size=10000)
In [8]: b = np.random.choice([True, False], size=10000)
In [9]: %timeit a ^ b
The slowest run took 7.61 times longer than the fastest. This could mean that an intermediate result is being cached
100000 loops, best of 3: 11 us per loop
In [10]: %timeit np.logical_xor(a,b)
The slowest run took 6.25 times longer than the fastest. This could mean that an intermediate result is being cached
100000 loops, best of 3: 11 us per loop
2。大小为 1000
的随机布尔序列
In [11]: a = np.random.choice([True, False], size=1000)
In [12]: b = np.random.choice([True, False], size=1000)
In [13]: %timeit a ^ b
The slowest run took 21.52 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 1.58 us per loop
In [14]: %timeit np.logical_xor(a,b)
The slowest run took 19.45 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 1.58 us per loop
3。大小为 100
的随机布尔序列
In [15]: a = np.random.choice([True, False], size=100)
In [16]: b = np.random.choice([True, False], size=100)
In [17]: %timeit a ^ b
The slowest run took 33.43 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 614 ns per loop
In [18]: %timeit np.logical_xor(a,b)
The slowest run took 45.49 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 616 ns per loop
4.大小为 10
的随机布尔序列
In [19]: a = np.random.choice([True, False], size=10)
In [20]: b = np.random.choice([True, False], size=10)
In [21]: %timeit a ^ b
The slowest run took 86.10 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 509 ns per loop
In [22]: %timeit np.logical_xor(a,b)
The slowest run took 40.94 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 511 ns per loop
我知道逻辑与是&，逻辑或是|在 Pandas 系列中，但我一直在寻找按元素逻辑异或。我想我可以用 AND 和 OR 来表达它，但如果可以的话，我更愿意使用 XOR。
谢谢！
Python异或：a ^ b
Numpy logical XOR: np.logical_xor(a,b)
测试性能 - 结果相等：
1.大小为 10000
的随机布尔序列In [7]: a = np.random.choice([True, False], size=10000)
In [8]: b = np.random.choice([True, False], size=10000)
In [9]: %timeit a ^ b
The slowest run took 7.61 times longer than the fastest. This could mean that an intermediate result is being cached
100000 loops, best of 3: 11 us per loop
In [10]: %timeit np.logical_xor(a,b)
The slowest run took 6.25 times longer than the fastest. This could mean that an intermediate result is being cached
100000 loops, best of 3: 11 us per loop
2。大小为 1000
的随机布尔序列In [11]: a = np.random.choice([True, False], size=1000)
In [12]: b = np.random.choice([True, False], size=1000)
In [13]: %timeit a ^ b
The slowest run took 21.52 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 1.58 us per loop
In [14]: %timeit np.logical_xor(a,b)
The slowest run took 19.45 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 1.58 us per loop
3。大小为 100
的随机布尔序列In [15]: a = np.random.choice([True, False], size=100)
In [16]: b = np.random.choice([True, False], size=100)
In [17]: %timeit a ^ b
The slowest run took 33.43 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 614 ns per loop
In [18]: %timeit np.logical_xor(a,b)
The slowest run took 45.49 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 616 ns per loop
4.大小为 10
的随机布尔序列In [19]: a = np.random.choice([True, False], size=10)
In [20]: b = np.random.choice([True, False], size=10)
In [21]: %timeit a ^ b
The slowest run took 86.10 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 509 ns per loop
In [22]: %timeit np.logical_xor(a,b)
The slowest run took 40.94 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 511 ns per loop